If displacement x and velocity v are related | vedclass

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(C) The given equation is $4v^2 = 25 - x^2$.Rearranging the terms,we get $x^2 + 4v^2 = 25$.Dividing both sides by $25$,we get $\frac{x^2}{25} + \frac{

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$4\pi$

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(C) The given equation is $4v^2 = 25 - x^2$.Rearranging the terms,we get $x^2 + 4v^2 = 25$.Dividing both sides by $25$,we get $\frac{x^2}{25} + \frac{4v^2}{25} = 1$,which can be written as $\frac{x^2}{5^2} + \frac{v^2}{(5/2)^2} = 1$.Comparing this with the standard equation of an ellipse for $SHM$,$\frac{x^2}{A^2} + \frac{v^2}{(A\omega)^2} = 1$,we get:Amplitude $A = 5$ and $A\omega = \frac{5}{2}$.Substituting $A = 5$ into the second expression: $5\omega = \frac{5}{2} \Rightarrow \omega = \frac{1}{2} 	ext{ rad/s}$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.Substituting $\omega = \frac{1}{2}$,we get $T = \frac{2\pi}{1/2} = 4\pi 	ext{ s}$.

If displacement $x$ and velocity $v$ are related as $4v^2 = 25 - x^2$ in a $SHM$. Then the time period of the given $SHM$ is (Consider $SI$ units).

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